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Table of Contents

 


 

Chi-Square Test of Independence

Used for comparing frequencies (counts) of nominal or ordinal level data for two samples across two or more subgroups displayed in a crosstabulation table.  More common and more flexible than z-tests of proportions.

 

Problem: You wish to evaluate the association between a person's sex and their attitudes toward school spending on athletic programs. A random sample of adults in your school district produced the following table (counts).

 

Female

Male

Row Total

Spend more money

15

25

40

Spend the same

5

15

20

Spend less money

35

10

45

Column Total

55

50

105

Assumptions

Independent random sampling

Nominal/Ordinal level data

No more than 20% of the cells have an expected frequency less than 5

No empty cells

 

State the Hypothesis

Ho: There is no association between a person's sex and their attitudes toward spending on athletic programs.

Ha: There is an association between a person's sex and their attitudes toward spending on athletic programs.

 

Set the Rejection Criteria

Determine degrees of freedom df=(# of rows - 1)(# of columns - 1)

df=(3 - 1)(2 - 1) or df=2

Establish the confidence level (.05, .01, etc.);   Alpha = .05

Based on the chi-square distribution table, the critical value = 5.991

 

Compute the Test Statistic

  where

Fo= observed frequency

Fe= expected frequency for each cell

Fe=(frequency for the column)(frequency for the row)/n

 

 

Frequency Observed

 

Female

Male

Row Total

Spend more

15

25

40

Spend same

5

15

20

Spend less

35

10

45

Column Total

55

50

105

 

 

Frequency Expected

 

Female

Male

Row Total

Spend more

55*40/105 = 20.952

50*40/105 = 19.048

40

Spend same

55*20/105 = 10.476

50*20/105 = 9.524

20

Spend less

55*45/105 = 23.571

50*45/105 = 21.429

45

Column Total

55

50

105

 

 

Chi-square Calculations

 

Female

Male

Spend more

(15-20.952)2/20.952

(25-19.048)2/19.048

Spend same

(5-10.476)2/10.476

(15-9.524)2/9.524

Spend less

(35-23.571)2/23.571

(10-21.429)2/21.429

 

 

Chi-square

 

Female

Male

Spend more

1.691

1.860

Spend same

2.862

3.149

Spend less

5.542

6.096

 

21.200

 

 

 

Decide Results of Null Hypothesis

Since the chi-square test statistic 21.2 exceeds the critical value of 5.991, you may conclude there is a statistically significant association between a person's sex and their attitudes toward spending on athletic programs. As is apparent in the contingency table, males are more likely to support spending than females.

Software Output Example


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